Do \(SA\perp\left(ABCD\right)\Rightarrow\widehat{SCA}\) là góc giữa SC và (ABCD)
\(\Rightarrow\widehat{SCA}=30^0\)
\(\Rightarrow SA=AC.tan30^0=\frac{a\sqrt{6}}{3}\)
\(S_{MAN}=\frac{1}{2}AM.AN=\frac{1}{2}.\frac{1}{2}a.\frac{1}{2}a=\frac{a^2}{8}\)
\(S_{CDN}=\frac{1}{2}CD.DN=\frac{1}{2}a.\frac{1}{2}a=\frac{a^2}{4}\)
\(\Rightarrow S_{MBCN}=S_{ABCD}-\left(S_{MAN}+S_{CDN}\right)=a^2-\frac{a^2}{8}-\frac{a^2}{12}=\frac{19}{24}a^2\)
\(\Rightarrow V_{S.MBCN}=\frac{1}{3}SA.S_{MBCN}=\frac{1}{3}.\frac{a\sqrt{6}}{3}.\frac{19}{24}a^2=\frac{19a^3\sqrt{6}}{216}\)